## quiz 1 fall 2014 prep #10

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Armo_Derbarsegian_3K
Posts: 35
Joined: Sat Jul 09, 2016 3:00 am

### quiz 1 fall 2014 prep #10

Hello,

I did #10 and I was looking at the answers in the back and noticed:

"USE BOHR EQU. FOR H ATOM (ANY 1e- H-LIKE ATOM, Li^2+):

En = -2.178x10^-18 J (Z^2/n^2) for n=1,2,3....

j=joules, z=atomic number, n=energy level"

What is the difference between that equation and using -hR? When do we need to plug in the Z and n?

I just wrote it as -7.55x10^-20J which I derived from the frequency given and set it equal to:

-7.55x10^-20J=-hR/4^2 -hR/n^2

-7.55x10^-20J=-hR(1/16-1/n^2)
----------------- ---
-hR -hR

and solved it that way. Is this work still valid? Is that what the answer key did and just didn't show the work? Where/when does the Z and n come into play?

Thanks!

Chem_Mod
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Joined: Thu Aug 04, 2011 1:53 pm
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### Re: quiz 1 fall 2014 prep #10

Ignore the above equation.

Here is the updated solution:

delta E = hv = (6.63 x 10-34J · s) (1.14 x 1014 s-1) = 7.56 x 10-20J

Since light is emitted delta E = - 7.56 x 10-20J

USE BOHR EQU. FOR H-ATOM:

En = -2.178 x 10 -18J ( 1/n2 ) for n = 1, 2, ... 

WHERE, J: JOULES n: ENERGY LEVEL

So E = E4 - En (because light is emitted)
= - 7.56 x 10-20 J = -2.178 x 10-18J (1/42 - 1/n2 )

- 7.56 x 10-20J -2.178 x 10-18J = (1/42 - 1/n2 )

3.47 x 10-2 = 1/16 - 1/n2

1/n2 = 2.78 x 10-2
n = 6

Josh_Zhong_1F
Posts: 34
Joined: Fri Sep 30, 2016 3:02 am

### Re: quiz 1 fall 2014 prep #10

Hi,

Where does the number -2.178x10^-18J come from??

Chem_Mod
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Joined: Thu Aug 04, 2011 1:53 pm
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### Re: quiz 1 fall 2014 prep #10

404412260 wrote:Hi,

Where does the number -2.178x10^-18J come from??

(Planck's constant)x(Rydberg constant)

Jamie Huang 1L
Posts: 20
Joined: Fri Jul 15, 2016 3:00 am

### Re: quiz 1 fall 2014 prep #10

For this question, I got 7.56 x 10^-20 J for when freq is equal to 1.14 x 10^14 Hz, then using n=4, I plugged in -hR/n^2 and I got -1.36 x 10 ^-19 J. When I try to subtract the Energy of n=4 from Energy of freq, I got -2.12 x 10^-19 J. Using this Energy, I try to find n by plugging values into the formula E= -hR/n^2, but somehow I got the answer 3, and the correct answer is 6. Is there anything that I did wrong in my work?